英文:
Why isn't this assignment a type mismatch?
问题
为什么这不是类型不匹配的情况？
来自：https://golang.org/ref/spec#Assignability
在以下任何情况下，值x都可以分配给类型为T的变量（"x可以分配给T"）：
...省略...
x的类型V和T具有相同的基础类型，并且V或T中至少有一个不是命名类型。
...省略...
这是因为N[]的基础类型是N[]，而N[]不是命名类型吗？
这背后的原理是什么？
package main
import "fmt"
type N []N
func main() {
n := make([]N, 1)
fmt.Printf("%T\n", n)
fmt.Printf("%T\n", n[0])
n[0] = n
//fmt.Println(n)
}
*输出：*
[]main.N
main.N
英文:
Why isn't this a type mismatch?
From: https://golang.org/ref/spec#Assignability
A value x is assignable to a variable of type T ("x is assignable to T") in any of these cases:
...snip...
x's type V and T have identical underlying types and at least one of V or T is not a named type.
...snip...
Is that because the underlying type of N[] is N[] which is not a named type?
What's the rationale behind it?
package main
import "fmt"
type N []N
func main() {
n := make([]N, 1)
fmt.Printf("%T\n", n)
fmt.Printf("%T\n", n[0])
n[0] = n
//fmt.Println(n)
}
*Output:*
[]main.N
main.N
答案1
得分: 1
你正在询问 n[0] = n 是否有效。你正确地从语言规范中找到了规则：
在以下情况下，值
x可以赋给类型为T的变量（“x可以赋给T”）：
- ...
x的类型V和T具有相同的基础类型，并且V或T中至少有一个不是命名类型。
以下是它在这里的应用：
n[0]的类型是N，基础类型是[]N（来自语言规范：“N在其类型声明中引用的类型”）。n的类型是[]N（具有相同的基础类型）。
因此，n[0] 和 n 具有相同的基础类型（[]N），且 n 的类型不是命名类型。
英文:
You're asking if n[0] = n is valid. You've correctly identified the rule from the language spec:
> A value x is assignable to a variable of type T ("x is
> assignable to T") in any of these cases:
>
> * ...
> * x's type V and T have identical underlying types and at least one of V or T is not a named type.
And here's how it applies here:
n[0]has typeNand underlying type[]N(from the language specification: "the type to whichNrefers in its type declaration").nhas type[]N(with the same underlying type).
So n[0] and n have identical underlying types ([]N), and the type of n is not a named type.
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